© Boardworks Ltd 20051 of 58 © Boardworks Ltd 2005 1 of 58 AS-Level Maths: Statistics 1 for Edexcel S1.4 Correlation and regression This icon indicates.

© Boardworks Ltd 20051 of 58 © Boardworks Ltd 20051 of 58

AS-Level Maths: Statistics 1for Edexcel

S1.4 Correlation and regression

This icon indicates the slide contains activities created in Flash. These activities are not editable.

For more detailed instructions, see the Getting Started presentation.

© Boardworks Ltd 20052 of 58

Co

nte

nts


Scatter graphs

Scatter graphs, types of correlation and lines of best fit

Product–moment correlation coefficients

The effects of coding on correlation

Regression


There are many situations where people wish to find out whether two (or more) variables are related to each other. Here are some examples:

Correlation

Is systolic blood pressure related to age?

Is the life expectancy of people in a country related to how wealthy the country is?

Are A-level results related to the number of hours students spend undertaking part-time work?

Is an athlete’s leg length related to the time in which they can run 100m?

Correlation is a measure of relationship – the stronger the correlation, the more closely related

the variables are likely to be.


Example: The table shows the latitude and mean January temperature (°C) for a sample of 10 cities in the northern hemisphere.

Scatter graphs are a useful visual way of judging whether a relationship appears to exist between two variables.

Scatter graphs

City Latitude Mean Jan. temp. (°C)

Belgrade 45 1

Bangkok 14 32

Cairo 30 14

Dublin 50 3

Havana 23 22

Kuala Lumpur 3 27

Madrid 40 5

New York 41 0

Reykjavik 30 –1

Tokyo 36 5


The data in the table can be presented in a scatter graph:

Scatter graph showing how January temperatures change with latitude

-5

0

5

10

15

20

25

30

35

0 10 20 30 40 50 60

Latitude

Tem

per

atu

re (

ºC)

Scatter graphs

This shows that mean January temperature tends to decrease as the latitude of the city increases. We say that the variables are negatively correlated.


In this example, a city’s temperature is likely to be dependent upon its latitude – not the other way around. Temperature cannot affect a city’s latitude.

The latitude is called the independent (or explanatory) variable. The temperature is called the dependent(or response) variable.

Scatter graphs

When plotting scatter graphs, the convention is to always plot the independent variable on the horizontal axis and the dependent variable on the vertical axis.

Scatter graph showing how January temperatures change with latitude

-5

0

5

10

15

20

25

30

35

0 10 20 30 40 50 60

Latitude

Tem

per

atu

re (

ºC)


Scatter graphs


The type of correlation existing between two variables can be described in terms of the gradient of the slope formed by the points, and how close the points lie to a straight line.

Strong positive correlation – the points lie close to a straight line with positive gradient.

Weak positive correlation – the points are more scattered but follow a general upward trend.

Correlation


Strong negative correlation – the points lie close to a straight line with negative gradient.

Weak negative correlation – the points are more scattered but follow a general downward trend.

Correlation

0

5

10

15

20

25

30

0 2 4 6 8 10 12x

y

0

5

10

15

20

25

30

0 2 4 6 8 10 12

x

y


No correlation –the points are scattered across the graph area indicating no relationship between the variables.

Correlation

0

5

10

15

20

25

0 2 4 6 8 10 12

x

y


The following diagram illustrates why it is important to interpret scatter diagrams with caution.

The diagram shows life expectancy at birth plotted against annual cigarette consumption for a sample of 9 countries.

Correlation vs. causation


The diagram shows a positive correlation between cigarette consumption and life expectancy. However, it would be wrong to conclude that consuming more cigarettes causes people to live longer.

Correlation vs. causation

This type of correlation is sometimes referred to asnonsense correlation.

The relationship can be explained because both life expectancy and cigarette consumption for a country are correlated with a third variable – the wealth of the country.


When a linear relationship exists between two variables,a line of best fit can be drawn on the scatter graph.

Lines of best fit


Lines of best fit

It can be shown that a line of best fit always passes through the mean point, .( , )x y

mean point

Example: A line of best fit can be added to the scatter graph showing mean January temperatures and latitude.


Lines of best fit

The line of best fit can be used to make predictions.

For example, Los Angeles has a latitude of 34°N. The line of best fit suggests that Los Angeles should have a January temperature of about 9°C.

The actual mean temperature for Los Angeles is 13°C.


Co

nte

nts



Product–moment correlation coefficient


Regression



The product–moment correlation coefficient (r) gives a numerical measure of the strength of the linear association between two variables.


This means that it measures how close the points on a scatter graph lie to a straight line.


The product–moment correlation coefficient works so that:


0

20

40

60

80

100

0 1 2 3 4

0

20

40

60

80

100

0 1 2 3 4

• r = 1 indicates perfect linear positive correlation;

0

20

40

60

80

100

0 1 2 3 4

–1 ≤ r ≤ 1

• r = –1 indicates perfect linear negative correlation;

• r = 0 indicates that there is absolutely no linear correlation between the variables.


i ii i

x yx y

n

i

i

xx

n

2

2

i

i

yy

n

2

2


The product–moment correlation coefficient for n pairs of observations is obtained using the formula:( , ), ..., ( , )n nx y x y1 1

xy

xx yy

Sr

S S

( )ix x 2

xxS

( )iy y 2

yyS

( )( )i ix x y y xySwhere:

Usually, the second version of each formula is used.


Example: The table shows the average body mass and brain mass of 6 species of animal.


Species Body mass (kg) Brain mass (g)

Baboon 11 180

Cat 3.3 30

Fox 4.2 50

Mouse 0.02 0.4

Monkey 10 120

Rabbit 2.5 12

a) Draw a scatter graph showing brain mass plotted against body mass.

b) Describe the relationship that exists between the two variables.

c) Calculate the product–moment correlation coefficient.



Scatter graph comparing body mass and brain mass

0

50

100

150

200

0 2 4 6 8 10 12

Body mass (kg)

Bra

in s

ize

(kg

)

a)

b) The scatter graph shows strong positive correlation, meaning that the size of an animal’s brain tends to increase as its body mass increases.



. ... . .11 3 3 2 5 31 02x ... .180 30 12 392 4y

. ... . .2 2 2 211 3 3 2 5 255 78x ...

2 2 2 2180 30 12 50344y ( ) ... ( . )11 180 2 5 12 3519xy

c) Species Body mass (kg) Brain mass (g)

Baboon 11 180

Cat 3.3 30

Fox 4.2 50

Mouse 0.02 0.4

Monkey 10 120

Rabbit 2.5 12



So: i ii i

x yx y

n xyS

i

i

yy

n

2

2yyS

i

i

xx

n

2

2xxS

xy

xx yy

Sr

S S Therefore:

Note: The product–moment correlation coefficient can be found using built-in calculator functions.

. .

31 02 392 43519

61490

..

231 02255 78

6.95 41

. 2

392 4

50 3446

24 681

.

1490

95 41 24 681

.0 971


Examination-style question: A researcher believes there is a relationship between a country’s annual income per head (x, in $1000) and the per capita carbon dioxide emissions (c, tonnes). He collects data from a random sample of 10 countries and records the following results:


.

. .

.

2 2

96 5 54

2156 9 383 54

619 6

x c

x c

xc

Calculate the value of the product–moment correlation coefficient and comment on the implications of your answer.


i

cc i

cS c

n

2

2

i

xx i

xS x

n

2

2

i ixc i i

x cS x c

n So:

Therefore, the product–moment correlation coefficient is:


xc

xx cc

Sr

S S

Income shows weak positive correlation with CO2 emissions – emissions are generally higher in wealthier countries. However, as the correlation is low, the result is somewhat inconclusive.

..

96 5 54619 6

10.98 5

..

296 52156 9

10.1225 675

. 254

383 5410

.91 94

.

. .

98 5

1225 675 91 94. (3 s0 293 .f.)


Co

nte

nts



Product-moment correlation coefficient


Regression

Effects of coding on correlation


The value of the product–moment correlation coefficient is unaffected by linear transformations of the variables.

More specifically, if the variables u and v are related to the variables x and y through the transformations

Effect of coding on the correlation

then the correlation coefficient between u and v is identical to the correlation coefficient between x and y.

v = cy + d

u = ax + b

Note: this is only true if c and a are greater than 0.




Example: The heights (in cm) of a sample of 11 men and their adult sons can be summarized as follows:

Solution: Let u = x – 160 and v = y – 160.

Then:

where x = height of father (in cm) and y = height of son (in cm).

Calculate the value of the product-moment correlation coefficient between the fathers’ and sons’ heights.


2 2

105 142 2483

1959 3326

u v uv

u v

( ) ( ) ( )( )

( ) ( )2 2

160 105 160 142 160 160 2483

160 1959 160 3326

x y x y

x y


uvS

105 142

248311

As the transformations between (x, y) and (u, v) are linear, the correlation coefficient between the fathers’ and sons’ heights must also be 0.943.

We can find the values of Suv, Suu and Svv:


.

. .r

1127 55

956 727 1492 909So:

.1127 55

21051959

11uuS .956 727

vvS 2142

332611

.1492 909

0.943 (to 3 s.f.)


The product–moment correlation coefficient (PMCC) measures the strength of a linear relationship.

However:

Limitations of the PMCC

0

5

10

15

20

0 2 4 6 8 100

5

10

15

20

25

0 2 4 6 8 10

Outliers can greatly distort the PMCC;

The PMCC is not a suitable measure of correlation if the relationship is non-linear.


Variables can be described as being either:

Types of variables

Sometimes a variable is controlled by the experimenter – they decide in advance what values that variable should take. If a variable is controlled, then it is non-random.

Random variables take values that cannot be predicted with certainty before collecting the data.

random or

non-random.


Example: An experiment is carried out into how fast a mug of coffee cools. The temperature of the coffee is measured every 2 minutes until 10 minutes have passed.

Types of variables

Time (minutes) 0 2 4 6 8 10

Temperature (°C) 95 83 73 64 55 48

The values for the time were chosen by the experimenter. If the experiment is repeated, the values for the time will be the same. Therefore, time is a non-random variable.

Temperature is a random variable. The values for this variable may be different if the experiment is repeated.


Co

nte

nts



Product-moment correlation coefficient


Regression

Regression


Linear regression involves finding the equation of the line of best fit on a scatter graph.

The equation obtained can then be used to make an estimate of one variable given the value of the other variable.

There are two cases to consider, depending upon whether:

Regression – random on random

We deal first with the situation where both variables (x and y) are random, and where we wish to predict a value for y given a value for x.

1. We wish to find a value of y given a value for x, or

2. We want to estimate x given y.



The best fitting line is the one that minimizes the sum of the squared deviations, , where di is the vertical distance between the ith point and the line.

2id

d1d2

d3

d4d5

d6

The distances di are sometimes referred

to as residuals.



As stated previously, the best fitting line should pass through the mean point, .( , )x y


The line that minimizes the sum of squared deviations is formally known as the least squares regression line of y on x.

The equation of the least squares regression line of y on x is:


2

2xx

xS x

n

and: a y bx

xy

x yS xy

n Recall: and

y = a + bx

b is sometimes referred to as the regression

coefficient.

xy

xx

Sb

Swhere:


Consider again the temperature data presented earlier.

Example: The table shows the latitude, x, and mean January temperature(°C), y, for a sample of 10 cities in the northern hemisphere.

Calculate the equation of the regression line of y on x and use it to predict the mean January temperature for the city of Los Angeles, which has a latitude of 34°N.


City Latitude Mean Jan. temp. (°C)

Belgrade 45 1

Bangkok 14 32

Cairo 30 14

Dublin 50 3

Havana 23 22

Kuala Lumpur 3 27

Madrid 40 5

New York 41 0

Reykjavik 30 –1

Tokyo 36 5


2 11 636x


We begin by finding summary statistics for the table:

x 312

We then use these to calculate the gradient (b) and y-intercept (a) for the regression line.

City Latitude (x)

Mean Jan. temp. (°C) (y)

Belgrade 45 1

Bangkok 14 32

Cairo 30 14

Dublin 50 3

Havana 23 22

Kuala Lumpur 3 27

Madrid 40 5

New York 41 0

Reykjavik 30 –1

Tokyo 36 5

y 108

y 2 2494

xy 2000



xy

x yS xy

n

2

2

312

108

11 636

2494

2000

x

y

x

y

xy

xx

xS x

n

2

2

To find the gradient, we need Sxy and Sxx:

Therefore:

xy

xx

Sb

S

312 1082000

10. 1369 6

2

312

11 63610

.1901 6

.

.

1369 6

1901 6–0.720 (to 3 s.f.)


Therefore, the equation of the regression line is:

y = 33.3 – 0.720x

This is our estimate of the mean January temperature in Los Angeles.


x 312

10

y 108

10

To find the y-intercept we also need and :x y

So: a y bx

.31 2

.10 8

. ( . . )10 8 0 720 31 2

= 33.3 (to 3 s.f.)

So, when x = 34, y = 33.3 – 0.720 × 34 = 8.82°C.

2

2

312

108

11 636

2494

2000

x

y

x

y

xy


This prediction for the mean January temperature in Los Angeles is based purely on the city’s latitude.

There are likely to be additional factors that can affect the climate of a city, for example:


The concept of regression we have considered here can be extended to incorporate other relevant factors, producing a new formula. This allows for more accurate prediction.

altitude;

proximity to the coast;

ocean currents;

prevailing winds.


A regression equation can only confidently be used to predict values of y that correspond to x values that lie within the range of the data values available.

The dangers of extrapolation

It can be dangerous to extrapolate (i.e. to predict) from the graph, a value for y that corresponds to a value of x that lies beyond the range of the values in the data set.

It is reasonably safe to make predictions

within the range of the data.

It is unwise to extrapolate beyond the given data.

This is because we cannot be sure that the relationship between the two variables will continue to be true.


Examination-style question: The average weight and wingspan of 9 species of British birds are given in the table.

Examination-style question: regression

Bird Weight (g)

Wingspan (cm)

Wren 10 15

Robin 18 21

Great tit 18 24

Cuckoo 57 33

Blackbird 100 37

Pigeon 300 67

Lapwing 220 70

Crow 500 99

Common gull 400 100

a) Plot the data on a scatter graph. Comment on the relationship between the variables.

b) Calculate the regression line of wingspan on weight.

c) Use your regression line to estimate the wingspan of a jay, if its average weight is 160 g.

d) Explain why it would be inappropriate to use your lineto estimate the wingspan of a duck, if the averageweight of a duck is 1 kg.



a)

The graph indicates that there is fairly strong positive correlation between weight and wingspan – this means that wingspan tends to be longer in heavier birds.


b) Summary values for the paired data are:


x 1623

xy

x yS xy

n

xx

xS x

n

2

2

xy

xx

Sb

S

These can be used to find the gradient of the regression line:

Therefore:

x = weighty = wingspany 466

2 562 397x 2

32 890y 131 541xy

1623 466

131 5419

47 505.672

1623

562 3979

269 716

47 505.67

269 7160.176 (to 3 s.f.)



.1623

180 339

x

.y 466

51 789

To find the y-intercept we also need and :x y

So: a y bx

Therefore, the equation of the regression line is:

y = 20.0 + 0.176x

where y = wingspan and x = weight.

. ( . . ) 51 78 0 176 180 33

.20 04


c) When the weight is 160 g, we can predict the wingspan to be:

y = 20.0 + 0.176x =

d) The average weight of a duck is outside the range of weights provided in the data. It would therefore be inappropriate to use the regression line to predict the wingspan of a duck, as we cannot be certain that the same relationship will continue to be true at higher weights.

Note: The regression coefficient (0.176) can be interpreted here as follows: as the weight increases by 1 g, the wingspan increases by 0.176 cm, on average.


20.0 + (0.176 × 160)= 48.2 cm (to 3 s.f.)


We now turn our attention to the situation where we wish to estimate a value of x when we are given a value of y. We will continue to assume that both variables are random.

To predict x given y (when both variables are random), we use the regression line of x on y. This line has the equation:

Predicting x from y – random on random

x = a′ + b′y

Note that both the regression line of x on y and the regression line of y on x pass through the mean point.

The two lines won’t in general be equal, unless the points lie in a perfect straight line.

xy

yy

Sb

S and a x by where:

This regression line is designed to minimize

the sum of the squares of the deviations in the x

direction.


Examination-style question: 15 AS-level mathematics students sit papers in C1 and S1. Their results are summarized below, with c representing the percentage mark in C1, and s the percentage mark in S1.


2 2 888 58 362 943 66 445 61 878c c s s cs

a) Calculate the regression line of s on c and the regression

line of c on s.b) Caroline was absent for her C1 examination, but scored

51% in S1. Use the appropriate regression line to estimate her percentage score in the C1 paper.c) Calculate the product–moment correlation coefficient between the marks in the two papers. Comment on theimplications of this for the accuracy of the estimate found in b).



2

888

58 36215ccS

a) From these summary values we can calculate:

For the regression line of s on c:

.

.b

6052 4

5792 4

Also:

. , .c s 59 2 62 8667

So, the equation of the regression line of s on c is:

2 2 888 58 362 943 66 445 61 878c c s s cs

.7161 7332

943

66 44515ssS .5792 4

888 943

61 87815csS

.6052 4

. ( . . )a 62 8667 1 0449 59 2.1 0449 .1 00862

s = 1.01 + 1.04c



For the regression line of c on s:.

.b

6052 4

7161 733

. .

.

2 2

888 94358 362 5792 4 66 445 7161 733

15 15888 943

61 878 6052 415

cc ss

cs

S S

S

. .c s 59 2 62 8667

So, the equation of the regression line of c on s is:

c = 6.07 + 0.845s

b) We wish to estimate the value of c when s = 51. Both variables are random, so we use the regression line of c on s:

. ( . . )a 59 2 0 8451 62 8667.0 8451 .6 07

6.07 + (0.845 × 51) = 49.2c = 6.07 + 0.845s =

So we estimate Caroline to have scored 49% in C1.



c) The PMCC is calculated as follows:

cs

cc ss

Sr

S S

The PMCC indicates that there is very strong positive correlation between the marks in C1 and S1 – the points on the scatter graph would lie very close to a straight line.

This suggests that the mark estimated in b) is likely to be fairly accurate.

. .

.

2 2

888 94358 362 5792 4 66 445 7161 733

15 15888 943

61 878 6052 415

cc ss

cs

S S

S

. .c s 59 2 62 8667

.

. .

6052 4

5792 4 7161 733.0 94


Regression – controlled variables

We will now consider a situation where one of the variables (here assumed to be x) is a controlled variable. This means that the values of x are fixed – they were decided upon when the experiment was planned.

If x is a controlled variable, the regression line of x on y does not have any statistical meaning, since the values of x are not random.

We consequently use only the regression line of y on x, whether we are estimating a y or an x value.



Examination-style question: An agricultural researcher wishes to explore how the yield of a crop is affected by the amount of fertilizer used. She designs an experiment in which she fertilizes a small plot of land with a pre-determined amount of fertilizer. She obtains the following results:

Amount of fertilizer (kg), x 2 4 6 8 10 12

Crop yield (kg), y 8.55 9.34 9.52 10.39 11.42 11.57

a) Calculate the regression line of y on x.

b) The regression line of x on y is: x = –23.8 + 3.04yUse the appropriate regression line to estimate how much fertilizer would be needed to achieve a crop yield of 10 kg. Explain how you decided which regression line to use.



. . .2 242 364 60 79 623 20 447 74x x y y xy Also: , .7 10 132x y

The gradient of the regression line is: b = 22.2 ÷ 70 =

From these we get: Sxx = 70 and Sxy = 22.2

a) Amount of fertilizer (kg), x 2 4 6 8 10 12

Crop yield (kg), y 8.55 9.34 9.52 10.39 11.42 11.57

0.317

7.91and the intercept is: a = 10.132 – (0.317 × 7) =

Therefore the regression line is y = 7.91 + 0.317x.



b) Since x is a controlled variable, only the regression line of y on x has meaning. Therefore, this equation should be used to estimate x when y = 10:

Note: The intercept (7.91) represents the crop yield that might be expected if no fertilizer were to be applied. The equation of the line also shows that increasing the amount of fertilizer by

1 kg increases the expected crop yield by 0.317 kg.

y = 7.91 + 0.317x

10 = 7.91 + 0.317x2.09 = 0.317x

x = 6.59

© Boardworks Ltd 20051 of 58 © Boardworks Ltd 2005 1 of 58 AS-Level Maths: Statistics 1 for Edexcel S1.4 Correlation and regression This icon indicates.

Documents

correlation slide

correlation regression

types of correlation

type of correlation

strong negative correlation

strong positive correlation

weak negative correlation

weak positive correlation